3.723 \(\int e^{-3 \coth ^{-1}(a x)} x \sqrt {c-a^2 c x^2} \, dx\)

Optimal. Leaf size=151 \[ \frac {x^2 \sqrt {c-a^2 c x^2}}{3 \sqrt {1-\frac {1}{a^2 x^2}}}-\frac {3 x \sqrt {c-a^2 c x^2}}{2 a \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {4 \sqrt {c-a^2 c x^2}}{a^2 \sqrt {1-\frac {1}{a^2 x^2}}}-\frac {4 \sqrt {c-a^2 c x^2} \log (a x+1)}{a^3 x \sqrt {1-\frac {1}{a^2 x^2}}} \]

[Out]

4*(-a^2*c*x^2+c)^(1/2)/a^2/(1-1/a^2/x^2)^(1/2)-3/2*x*(-a^2*c*x^2+c)^(1/2)/a/(1-1/a^2/x^2)^(1/2)+1/3*x^2*(-a^2*
c*x^2+c)^(1/2)/(1-1/a^2/x^2)^(1/2)-4*ln(a*x+1)*(-a^2*c*x^2+c)^(1/2)/a^3/x/(1-1/a^2/x^2)^(1/2)

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Rubi [A]  time = 0.21, antiderivative size = 151, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {6192, 6193, 77} \[ \frac {x^2 \sqrt {c-a^2 c x^2}}{3 \sqrt {1-\frac {1}{a^2 x^2}}}-\frac {3 x \sqrt {c-a^2 c x^2}}{2 a \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {4 \sqrt {c-a^2 c x^2}}{a^2 \sqrt {1-\frac {1}{a^2 x^2}}}-\frac {4 \sqrt {c-a^2 c x^2} \log (a x+1)}{a^3 x \sqrt {1-\frac {1}{a^2 x^2}}} \]

Antiderivative was successfully verified.

[In]

Int[(x*Sqrt[c - a^2*c*x^2])/E^(3*ArcCoth[a*x]),x]

[Out]

(4*Sqrt[c - a^2*c*x^2])/(a^2*Sqrt[1 - 1/(a^2*x^2)]) - (3*x*Sqrt[c - a^2*c*x^2])/(2*a*Sqrt[1 - 1/(a^2*x^2)]) +
(x^2*Sqrt[c - a^2*c*x^2])/(3*Sqrt[1 - 1/(a^2*x^2)]) - (4*Sqrt[c - a^2*c*x^2]*Log[1 + a*x])/(a^3*Sqrt[1 - 1/(a^
2*x^2)]*x)

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 6192

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(c + d*x^2)^p/(x^(2*p)*(
1 - 1/(a^2*x^2))^p), Int[u*x^(2*p)*(1 - 1/(a^2*x^2))^p*E^(n*ArcCoth[a*x]), x], x] /; FreeQ[{a, c, d, n, p}, x]
&& EqQ[a^2*c + d, 0] &&  !IntegerQ[n/2] &&  !IntegerQ[p]

Rule 6193

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_)^2)^(p_.), x_Symbol] :> Dist[c^p/a^(2*p), Int[(u*(-1
+ a*x)^(p - n/2)*(1 + a*x)^(p + n/2))/x^(2*p), x], x] /; FreeQ[{a, c, d, n, p}, x] && EqQ[c + a^2*d, 0] &&  !
IntegerQ[n/2] && (IntegerQ[p] || GtQ[c, 0]) && IntegersQ[2*p, p + n/2]

Rubi steps

\begin {align*} \int e^{-3 \coth ^{-1}(a x)} x \sqrt {c-a^2 c x^2} \, dx &=\frac {\sqrt {c-a^2 c x^2} \int e^{-3 \coth ^{-1}(a x)} \sqrt {1-\frac {1}{a^2 x^2}} x^2 \, dx}{\sqrt {1-\frac {1}{a^2 x^2}} x}\\ &=\frac {\sqrt {c-a^2 c x^2} \int \frac {x (-1+a x)^2}{1+a x} \, dx}{a \sqrt {1-\frac {1}{a^2 x^2}} x}\\ &=\frac {\sqrt {c-a^2 c x^2} \int \left (\frac {4}{a}-3 x+a x^2-\frac {4}{a (1+a x)}\right ) \, dx}{a \sqrt {1-\frac {1}{a^2 x^2}} x}\\ &=\frac {4 \sqrt {c-a^2 c x^2}}{a^2 \sqrt {1-\frac {1}{a^2 x^2}}}-\frac {3 x \sqrt {c-a^2 c x^2}}{2 a \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {x^2 \sqrt {c-a^2 c x^2}}{3 \sqrt {1-\frac {1}{a^2 x^2}}}-\frac {4 \sqrt {c-a^2 c x^2} \log (1+a x)}{a^3 \sqrt {1-\frac {1}{a^2 x^2}} x}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 65, normalized size = 0.43 \[ \frac {\sqrt {c-a^2 c x^2} \left (a x \left (2 a^2 x^2-9 a x+24\right )-24 \log (a x+1)\right )}{6 a^3 x \sqrt {1-\frac {1}{a^2 x^2}}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x*Sqrt[c - a^2*c*x^2])/E^(3*ArcCoth[a*x]),x]

[Out]

(Sqrt[c - a^2*c*x^2]*(a*x*(24 - 9*a*x + 2*a^2*x^2) - 24*Log[1 + a*x]))/(6*a^3*Sqrt[1 - 1/(a^2*x^2)]*x)

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fricas [A]  time = 0.58, size = 42, normalized size = 0.28 \[ \frac {{\left (2 \, a^{3} x^{3} - 9 \, a^{2} x^{2} + 24 \, a x - 24 \, \log \left (a x + 1\right )\right )} \sqrt {-a^{2} c}}{6 \, a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(-a^2*c*x^2+c)^(1/2)*((a*x-1)/(a*x+1))^(3/2),x, algorithm="fricas")

[Out]

1/6*(2*a^3*x^3 - 9*a^2*x^2 + 24*a*x - 24*log(a*x + 1))*sqrt(-a^2*c)/a^3

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {-a^{2} c x^{2} + c} x \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(-a^2*c*x^2+c)^(1/2)*((a*x-1)/(a*x+1))^(3/2),x, algorithm="giac")

[Out]

integrate(sqrt(-a^2*c*x^2 + c)*x*((a*x - 1)/(a*x + 1))^(3/2), x)

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maple [A]  time = 0.05, size = 76, normalized size = 0.50 \[ -\frac {\left (-2 x^{3} a^{3}+9 a^{2} x^{2}-24 a x +24 \ln \left (a x +1\right )\right ) \sqrt {-c \left (a^{2} x^{2}-1\right )}\, \left (a x +1\right ) \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}}}{6 a^{2} \left (a x -1\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(-a^2*c*x^2+c)^(1/2)*((a*x-1)/(a*x+1))^(3/2),x)

[Out]

-1/6*(-2*x^3*a^3+9*a^2*x^2-24*a*x+24*ln(a*x+1))*(-c*(a^2*x^2-1))^(1/2)*(a*x+1)*((a*x-1)/(a*x+1))^(3/2)/a^2/(a*
x-1)^2

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {-a^{2} c x^{2} + c} x \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(-a^2*c*x^2+c)^(1/2)*((a*x-1)/(a*x+1))^(3/2),x, algorithm="maxima")

[Out]

integrate(sqrt(-a^2*c*x^2 + c)*x*((a*x - 1)/(a*x + 1))^(3/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x\,\sqrt {c-a^2\,c\,x^2}\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(c - a^2*c*x^2)^(1/2)*((a*x - 1)/(a*x + 1))^(3/2),x)

[Out]

int(x*(c - a^2*c*x^2)^(1/2)*((a*x - 1)/(a*x + 1))^(3/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(-a**2*c*x**2+c)**(1/2)*((a*x-1)/(a*x+1))**(3/2),x)

[Out]

Timed out

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